Question

When an inductor of inductance $\frac{6}{\pi }H,a$ capacitor of capacitance $\frac{50}{\pi }\mu F$ and resistor of resistance $R$ are connected in series with an AC supply of rms voltage $220V$ and frequency $50Hz$, the rms current through the circuit is $440mA$. Match the inductive reactance, $X_L$ the capacitive reactance, $X_C$ the resistance $R$ and the impedance $Z$ of the circuit given in List-I with the corresponding values given in List-II.

List I		List II
(A)	$X_L$	(I)	200 $\Omega$
(B)	$X_C$	(II)	300 $\Omega$
(C)	$R$	(III)	500 $\Omega$
(D)	$Z$	(IV)	600 $\Omega$

• A. A $\to$ IV , B $\to$ II, C $\to$ I , D $\to$ III
• B. A $\to$ IV , B $\to$ III, C $\to$ I , D $\to$ II
• C. A $\to$ IV , B $\to$ I, C $\to$ II , D $\to$ III
• D. A $\to$ I , B $\to$ IV, C $\to$ III , D $\to$ II

Answer 1

Answer: (C)

Given, inductance of inductor, $L=\frac{6}{\pi }H$
capacitance of capacitor, $C=\frac{50}{\pi }\mu F$
supply voltage, $V_{rms}=220V$,
supply frequency, $f=50Hz$
and supply current, $I_{rms}=440mA$
Now, (A) inductive reactance, $X_L=\omega L$
$X_L=2\pi f\times L$
$(\because \omega =2\pi f)$
$=2\pi \times 50\times \frac{6}{\pi }=600\Omega$
$X_L=600\Omega$
(B) Capacitance reactance, $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$
$X_C=\frac{\pi \times 10^6}{2\pi \times 50\times 50}$
$\Rightarrow X_C=\frac{10^6}{2500\times 2}$
$\Rightarrow X_C=200\Omega$
(C) Resistance, $R$
$\therefore$ Impedance of $LCR$ circuit,
$Z^2=R^2+{\left(X_L-X_C\right)}^2$
Putting the given values, we get
$(500{)}^2=R^2+(600-200{)}^2$
$250000=R^2+(400{)}^2$
$R^2=250000-160000$
$R^2=90000$
$\Rightarrow R=300\Omega$
(D) Impedance,
$Z=\frac{V_{rms}}{I_{rms}}=\frac{220}{440}\times 10^3$
$\Rightarrow Z=500\Omega$